In Problems 53–56, write a system of linear inequalities for the given graph.

The graph is in the first quadrant so the values are$x\ge 0$&$y\ge 0$.

The lines parallel to x-axis is$y=k$

here $k=8$then the equation is$y=8$

It include origin the inequality is$y\le 8$

The lines parallel to y-axis is$x=k$

here$k=5$ then the equation is$x=5$

it includes the origin the inequality is$x\le 5$

The line equation is $y-y_1=m(x-x_1)$

Here m=slope

$m=(\raisebox{1ex}{$y_2-y_1$}\!\left/ \!\raisebox{-1ex}{$(x_2-x1)$}\right.$

Here there are two lines the points for equation 1 is $(4,0)$& $\left(0,3\right)$.

Slope:localid="1646942857651" $m=\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

The line equation 1 for given points:

$$\begin{gathered} y-0=-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.(x-4) \\ 4y=-3x+12 \\ 3x+4y=12 \end{gathered}$$

Since the shaded region above the line the inequality is $3x+4y\ge 12$.

the given points are $(4,0)$&$(5,2)$

slope role="math" localid="1646943268903" $$\begin{gathered} m=\raisebox{1ex}{$(2-0)$}\!\left/ \!\raisebox{-1ex}{$(5-4)$}\right. \\ m=-2 \end{gathered}$$

the line equation ismrole="math" localid="1646943255354" $$\begin{gathered} y-0=2(x-4) \\ y=2x-8 \\ 2x-y=8 \end{gathered}$$

the shaded region is above the line so the inequality is$2x-y\le 8$

The inequalities of the shaded region is

$\left\{\begin{array}{l}x\ge 0\\ y\ge 0\\ 2x-y\le 8\\ 3x+4y\ge 12\\ x\le 5\\ y\le 8\end{array}\right.$